IP' be injective and define bn = aa(n) (in other words, the n'th term of the new sequence is the a(n)'th term of the old sequence).

Exercises 1. Let A and B be nonempty sets of rational numbers, with Au B = Q , such that a < b for all a E A, b E B. } (ii) If "( is rational, then exactly one of the conditions (*) or (**) holds. (iii) If "( is irrational, then both of the conditions (*) and (**) hold, and they may both be expressed as A = {r E Q: r < "( }, B = {r E Q: r > "( } . 2. With notations as in Exercise 1, the pair (A, B) is called a (Dedekind) cut of the rational field Q. (i) Show that the formulas A = {r E Q: r:::; O} U {r E Q: r> 0 and r2 < 2} B = {r E Q: r > 0 and r2 > 2 } define a cut (A, B) of the rationals.

Example. For each positive integer n, let Sn be a statement (which may be either true or false). Let A = {n E lID: Sn is true} . We say that Sn is true frequently if n E A frequently, and that Sn is true ultimately if n E A ultimately. ). 4. Theorem. 1, one and only one of the following conditions holds: (1) Xn E A ultimately; (2) Xn ¢. A frequently. Proof To say that (1) is false means that, whatever index N is proposed, the implication n ~ N :::} Xn E A 3. Sequences 36 is false, so there must exist an index n::::: N for which Xn ¢:.

### A First Course in Real Analysis (Undergraduate Texts in Mathematics) by Sterling K. Berberian

by Donald

4.2